Bi-elliptic Weierstrass points on curves of genus 5

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1 Bi-elliptic Weierstrass points on curves of genus 5 T. Kato, K. Magaard and H. Völklein A curve of genus > 5 has at most one bi-elliptic involution (i.e., involution yielding a quotient curve of genus 1), see [ACGH], VIII, C-2. This is an analogue of the well-known fact that a hyperelliptic curve of genus > 1 has only one hyperelliptic involution. Thus the geometry of bi-elliptic involutions is most interesting for curves C 5 of genus 5. A bi-elliptic involution of C 5 has exactly 8 fixed points. In the analogy with hyperelliptic involutions, these 8-sets correspond to the 4-sets of 2-division points of elliptic curves. Each fixed point of a bi-elliptic involution of C 5 is a Weierstrass point of weight 3 or 5. We call it a bi-elliptic Weierstrass point. A recent result of the first author [Ka2] says that conversely, each Weierstrass point of C 5 with gap sequence 1,2,3,5,9 or 1,2,3,5,7 is bi-elliptic. The number b of bi-elliptic involutions of C 5 satisfies b 5 and b 4. The genus 5 curves with b = 5 (the so-called Humbert curves) form a 2-dimensional family; each such curve has exactly 40 Weierstrass points and they all have weight 3 (see [Va] and Remark 2.8). In this paper we are mainly interested in the case b = 3. The Wiman curve (cf. [Wiman]) is the only genus 5 curve with the maximum number of 192 automorphisms, see Corollary 3.6(b) and Remark 3.7 below. It is also the only non-hyperelliptic genus 5 curve with the minimum number of 24 Weierstrass points, a recent result of Keem and Martens [KM]. This result is based on our corrected version of the paper [dc]: We show that the Wiman curve is the only genus 5 curve admitting three distinct bi-elliptic involutions whose fixed points are Weierstrass points of weight 5 (see Theorem 3.5). The paper [dc] contains the erroneous claim (Corollary 4.1 of [dc]) that there are 3 non-isomorphic curves with this property. The construction used to prove the uniqueness of the Wiman curve applies in a wider context, yielding a relation between the genus 5 curves admitting three bi-elliptic involutions and the genus 3 curves admitting three bi-elliptic involutions with product 1. Section 2.3 shows how the weight of the bielliptic Weierstrass points behaves under this correspondence. This is used for the above characterization of the Wiman curve (Theorem 3.5) and more generally, to show that the number of bi-elliptic Weierstrass points of weight 5 of a genus 5 field is 0,4,8 or 24 (Theorem 4.8). 1

2 There are many relations between Weierstrass points and automorphisms of a curve (e.g., the Lemma of Schoeneberg/Lewittes, see [Sch], [Le], [Towse], [MV]). For genus 5, Theorem 4.8(iii) shows that the property that all fixed points of a bi-elliptic involution α i have weight 5 is equivalent to the existence of an automorphism with square α i having 4 fixed points. In section 5 we classify the genus 5 curves whose automorphism group is transitive on the Weierstrass points. Acknowledgement This paper originated from a question of Gerriet Martens who doubted that Corollary 4.1 of [dc] was formulated correctly. We thank him for many valuable discussions and suggestions. Notation and terminology: We consider function fields of one variable over an algebraically closed field k of characteristic 0. We call them just function fields. By place of a function field we mean a place over k. By isomorphism of function fields we mean a k-isomorphism (i.e., a k-linear isomorphism). The rational function field generated by x is denoted by k(x). 1 Preliminaries 1.1 Weierstrass points Let X = X g be a function field of genus g 2. Let P be a place of X. An integer n 1 is called a gap at P if there is no f X having a pole of order n at P and no other pole; equivalently, if there is a regular differential form on X vanishing at P of order n 1. There are exactly g distinct gaps 1 = n 1 <... < n g < 2g at P, which form the gap sequence. The weight of P is w(p ) = g i=1 (n i i). The place P is called a Weierstrass point if w(p ) > 0 (equivalently, if n g > g). We denote the set of Weierstrass points of X by W(X). The weight of a Weierstrass point is an integer between 1 and g(g 1)/2, where the upper bound is attained only for hyperelliptic fields. The weighted number of Weierstrass points is g(g 1)(g + 1). For all these facts on Weierstrass points, see e.g. [ACGH], I.E. 1.2 Automorphisms of X and their fixed points For an automorphism α of X we let F X (α) denote the set of places of X fixed by α. The elements of F X (α) will be called the fixed points of α. We 2

3 call α fixed-point-free if F X (α) =. Let G be a (finite) group of automorphisms of X. For each place P of X, the stabilizer G P (i.e., the group of all α G with P F X (α)) is cyclic. This implies in particular: Lemma 1.1 P is fixed by at most one involutory automorphism of X. It is well-known that G has only finitely many non-regular orbits B 1,..., B r on the set of places of X. Let G i be the (cyclic) stabilizer in G of a point in B i (i = 1,..., r). Let d i be the order of G i. Let g 0 denote the genus of the fixed field X G. The Riemann-Hurwitz formula, applied to the extension X/X G, yields: 2(g 1) = 2 G (g 0 1) + G If G is an elementary abelian 2-group, this can be written as 2(g 1) = 2 G (g 0 1) + r (1 1 ) (1) d i i=1 ɛ G\{id} F X (ɛ) (2) A combinatorial analysis of equation (1) yields (see [Br], Lemma 3.18): If G 24(g 1), then r = 3, X G has genus 0 and one of the following holds: (L1) G = 84(g 1), {d 1, d 2, d 3 } = {2, 3, 7} (L2) G = 48(g 1), {d 1, d 2, d 3 } = {2, 3, 8} (L3) G = 40(g 1), {d 1, d 2, d 3 } = {2, 4, 5} (L4) G = 36(g 1), {d 1, d 2, d 3 } = {2, 3, 9} (L5) G = 30(g 1), {d 1, d 2, d 3 } = {2, 3, 10} (L6) G = (132/5)(g 1), {d 1, d 2, d 3 } = {2, 3, 11} (L7) G = 24(g 1), {d 1, d 2, d 3 } = {2, 3, 12} (L8) G = 24(g 1), {d 1, d 2, d 3 } = {2, 4, 6} (L9) G = 24(g 1), {d 1, d 2, d 3 } = {3, 3, 4} We call an involutory automorphism α of X a bi-elliptic involution if the fixed field X α has genus 1. A Weierstrass point of X is called bi-elliptic if it is fixed by a bi-elliptic involution. 3

4 2 Bi-elliptic involutions of genus 5 fields In this section we assume that X has genus g = Some basic results This subsection summarizes some basic results which are more or less folklore (cf. [Acc1], [Acc2], [KM]). For the convenience of the reader, we supply the simple proofs. Remark 2.1 Let E be a function field of genus 1. Then for each place Q of E and for each d {2, 3}, there is a rational subfield k(x) of E of degree d such that Q is totally ramified over this subfield. Since the automorphism group of E acts transitively on the places of E, it suffices to prove this for some Q. The case d = 2 is settled by the usual representation of E as a quadratic extension of k(x). For d = 3 note that E is the function field of a non-singular plane cubic, which allows us to use projection from a point in the plane not lying on the cubic, but lying on an inflection tangent. Lemma 2.2 Let α be a bi-elliptic involution of X. Then α has exactly 8 fixed points, and they are Weierstrass points of X of weight 3 or 5 with associated gap sequences 1,2,3,5,7, resp., 1,2,3,5,9. In particular, X is not hyperelliptic. Proof: We have F X (α) = 8 by (2). Let P F X (α), lying over the place Q of E := X α. Remark 2.1 yields a rational subfield k(x) of X of degree 4 resp.,6, such that P is totally ramified over k(x). Thus the gap sequence at P doesn t contain 4 and 6. This implies that P is a Weierstrass point of X. Hence X is not hyperelliptic (otherwise its Weierstrass points are the fixed points of the unique involution with rational fixed field, contradicting Lemma 1.1). Thus the gap sequence at P is of the form 1 = n 1 < 2 = n 2 < n 3 < n 4 < n 5 < 10. Since X is not hyperelliptic, P has weight w 5 by [Ka1]. If n 3 > 3 then n 3 5 and n 4 7, contradicting w 5. Thus n 3 = 3. If n 4 > 5 then n 4 7, contradicting w 5. Thus n 4 = 5. The claim follows. (See Lemma 3.1 of [KM] for a different argument). 4

5 Lemma 2.3 Let α 1, α 2 be two distinct bi-elliptic involutions of X. Then their product α 1 α 2 is a fixed-point-free involution whose fixed field is hyperelliptic of genus 3. Thus α 1, α 2 generate a Klein four group whose fixed field has genus 0. Proof: α 1, α 2 generate a dihedral group D of order 2l, where l 2. Then D has l involutions conjugate to α 1 or α 2. These l involutions are therefore bi-elliptic, hence the union of their fixed points is a set of 8l distinct places of X (by Lemma 1.1). The Riemann-Hurwitz formula, applied to the extension X/X D, yields that 8 = 2(g 1) 4l(g 0 1) + 8l 4l (3) where g 0 is the genus of X D. Hence l = 2, g 0 = 0, and the inequalities in (3) are equalities. Thus D is a Klein four group and the involution β := α 1 α 2 is fixed-point-free, hence X β has genus 3 by (2). Since [X β : X D ] = 2, it follows that X β is hyperelliptic. Lemma 2.4 Let α 1, α 2, α 3 be three distinct bi-elliptic involutions of X. Then the following holds: (i) The group A generated by α 1, α 2, α 3 is elementary abelian of order 8, and the field X A has genus 0. (ii) Each involution of A distinct from α 1, α 2, α 3 is fixed-point-free, hence its fixed field has genus 3. (iii) Let δ := α 1 α 2 α 3. Then each α i induces a bi-elliptic involution of the genus 3 field X := X δ. If P F X (α 1 ) lies over a Weierstrass point P of X, then P has weight 2. (iv) Let δ be the automorphism of E := X α 1 induced by δ. Then δ lies in the 2-torsion subgroup E[2] of the translation group of E. (v) Let P F X (α 1 ) and Q := δ(p ). Then the divisors 4P and 4Q are linearly equivalent on X. Proof: (i) A is elementary abelian of order 8 by Lemma 2.3. The fixed field of α 1, α 2 has genus 0 by Lemma 2.3, hence the same holds for X A. (ii) This follows because we have F X (ɛ) = 24 ɛ A\{id} 5

6 by (2). (iii) The fixed field of the group α i, δ has genus 1 by (2). This proves the first claim in (iii). Assume now that P F X (α 1 ) lies over a Weierstrass point P of X. Since P is fixed by the bi-elliptic involution induced by α 1, it follows that X is not hyperelliptic. Thus P has weight 2 if and only if 4 is a non-gap at P. If 4 is a gap at P then there is a regular differential form on X vanishing at P of order 3. This form pulls back to a regular differential form on X vanishing at P of order 3 (since X/ X is unramified). Thus 4 is a gap at P, contradicting Lemma 2.2. This completes the proof of (iii). (iv) Since δ has order 2, it suffices to prove that δ fixes no place P of E. So assume δ fixes such P. Let P be a place of X lying over P. Then δ(p ) = P or δ(p ) = α 1 (P ), hence P F X (δ) = or P F X (α 2 α 3 ) =. This contradiction proves (iv). (v) It follows from (iv) that δ commutes with each involution ι of E with rational fixed field k(x). There is one (and only one) such ι fixing the place P of E lying under P. Then ι also fixes Q := δ (P ). Thus X/k(x) is totally ramified at P and Q, hence for suitable choice of x we get that x is a function with divisor 4P 4Q. Remark 2.5 For each finite set P of places of k(x) of even cardinality there is a unique quadratic extension K P of k(x) (inside some fixed algebraic closure of k(x)) ramified exactly at the places in P. This follows from Riemann s existence theorem, see e.g. [V1], Th Lemma 2.6 Let α 1, α 2, α 3, α 4 be four distinct bi-elliptic involutions of X. Then their product α 5 is again a bi-elliptic involution (distinct from α 1, α 2, α 3, α 4 ), and X has exactly 5 bi-elliptic involutions. Moreover, the following holds: (i) The group H := α 1,..., α 5 is elementary abelian of order 16 and all involutions in H distinct from α 1,..., α 5 are fixed-point-free. (ii) The extension X/X H is ramified at exactly 5 places of the genus 0 field L := X H and equals the compositum of all quadratic extensions of L unramified outside these 5 places. (iii) H is normal in G := Aut(X) and G/H embeds as a subgroup of Aut(L) = PGL(2, k) permuting the 5 places ramified in X. Moreover, G/H permutes these 5 places transitively if and only if G = 160, which happens if and only if X is isomorphic to the extension of k(x) generated by the (x ζ)/(x ζ )), where ζ, ζ k are 5-th roots of unity. 6

7 Proof: It follows from Lemma 2.4 that H is elementary abelian of order 16 and each involution β H with β α 1 α 2 α 3 α 4 is fixed-point-free. Since F X (ɛ) = 40 ɛ H\{id} by (2), it follows that α 1 α 2 α 3 α 4 is bi-elliptic. Since the 8 fixed points of each α i are Weierstrass points of weight 3 and the weighted number of Weierstrass points of X is 120, it follows by Lemma 1.1 that X has at most 5 bi-elliptic involutions. Hence X has exactly 5 bi-elliptic involutions. For (ii), note that H induces on each F X (α i ) an abelian permutation group of order 8, hence acts transitively on F X (α i ). Hence the F X (α i ) are exactly the non-regular orbits of H on the set of places of X. Thus X/X H is ramified at exactly 5 places of L. The quadratic extensions of L inside X are the fixed fields of the 15 subgroups of index 2 of H, hence comprise all quadratic extensions of L unramified outside these 5 places by Remark 2.5. This proves (ii). If G/H has an element of order 5, then it permutes the 5 places transitively (because it cannot fix 5 places of L). Conversely, if G/H permutes the 5 places transitively, then they form an orbit of a cyclic subgroup of order 5 of PGL(2, k), hence they can be identified with the 5-th roots of unity for suitable choice of coordinates. In this case, G/H is the dihedral group of order 10. This proves (iii). Definition 2.7 A genus 5 field is called a Humbert field if it has (exactly) 5 bi-elliptic involutions (cf. [Va]). Remark 2.8 Each Humbert field can be written as k( x, x 1, x a, x b) with a b in k \{0, 1}. Thus the Humbert fields form a 2-dimensional family (cf. [Va]). This follows by putting three of the 5 places from (ii) to 0, 1,. Remark 2.9 (The automorphism group of a Humbert field) If the group G/H from (iii) is not transitive on the 5 places, then it is cyclic of order 2 or 4 or = S 3. Then G has order 16, 32, 64 or 96. 7

8 2.2 Genus 5 fields with more than 120 automorphisms Remark 2.10 The genus 5 field X has no automorphism of order 9. Each automorphism of X of order 5 has exactly 2 fixed points. Each automorphism of X of order 3 has exactly 4 or 7 fixed points. This follows from (1) and the structure of the fundamental group of a punctured surface, see [Br], Cor Proposition 2.11 Assume G = Aut(X) has order > 120. Then G acts transitively on W(X). Each Weierstrass point of X has the same weight w and is fixed by a bi-elliptic involution of X. Moreover, one of the following holds: (i) G = 192, w = 5 and G contains exactly 3 bi-elliptic involutions. (ii) G = 160, w = 3 and G contains exactly 5 bi-elliptic involutions. In case (ii), X is isomorphic to the (unique) Humbert field with 160 automorphisms (cf. Lemma 2.6(iii)). Proof: If G has order > 120 = 30(g 1), then we are in one of the cases (L1) (L4) from section 1.2. There is no Hurwitz curve of genus 5 (see [Co]), hence case (L1) does not occur here. Also case (L4) cannot occur by Remark It remains to consider the following two cases: Case (L2): Then G has exactly 3 non-regular orbits on the set of places of X of lengths 96, 64 and 24. The smallest orbit has length 24, hence W(X) 24 and so X is not hyperelliptic. Thus the number of fixed points of each involution in G is 0,4 or 8. Let τ be the element of order 8 of G that fixes a given point in the orbit of length 24. Let r be the number of fixed points of the involution τ 4. All places of X that ramify over X τ are fixed by τ 4, hence X/X τ is ramified at exactly r places of X. If r = 4 then either all of these 4 places are fixed by τ, or two are fixed by τ and the other two by τ 2, but not τ. Applying the Riemann-Hurwitz formula yields a contradiction in both cases (non-integer genus of X τ ). Thus r = 8, i.e., τ 4 is a bi-elliptic involution. Then each point in the orbit of length 24 is fixed by a bi-elliptic involution, hence has weight 3. These 24 points of weight 3 yield a count of at least 72 towards the weighted number of Weierstrass points which totals 120. Since W(X) is a union of G-orbits, these 24 points actually have weight 5 and comprise all of 8

9 W(X). Each of them is fixed by a bi-elliptic involution. Hence G has exactly 3 bi-elliptic involutions. Case (L3): Then G has exactly 3 non-regular orbits on the set of places of X of lengths 80,40 and 32. Moreover, L := X G has genus 0. Since G is solvable of order 160, it has a normal subgroup H of index 2 or 5. If [G : H] = 5 then X H is a non-trivial extension of the genus 0 field L ramified at only one place of L. This contradiction shows that [G : H] = 2. The G-orbit of length 32 splits into two H-orbits B, B of length 16. Each element τ H of order 5 fixes at least one point of B and one of B. Hence τ fixes exactly one point of B by Remark Thus H acts as a Frobenius group on B. The corresponding Frobenius kernel (whose elements 1 are the elements of H fixing no point of B) is then a (normal) Sylow 2-subgroup S of H, and C S (τ) = 1. The latter implies that τ does not normalize a proper non-trivial subgroup of S. Hence S is (elementary) abelian. Now let σ be an element of order 4 of G fixing a point in the G-orbit B of length 40. Then the involution α = σ 2 lies in H, hence S. Thus C H (α) = S, hence α has 8 fixed points on B and is therefore bi-elliptic. Then each point in B is fixed by a bi-elliptic involution, hence is a Weierstrass point of weight 3. Thus (ii) holds. The last claim in the Proposition follows from Lemma 2.6(iii). 2.3 A criterion for weight 5 A bi-elliptic Weierstrass point has weight w = 3 or w = 5 (by Lemma 2.2). The following Proposition shows that w is determined by the image of P in the genus 3 field X if X has at least 3 bi-elliptic involutions. Proposition 2.12 Let α 1, α 2, α 3 be three distinct bi-elliptic involutions of X. Let δ := α 1 α 2 α 3, P F X (α 1 ) and Q := δ(p ). Furthermore, let X := X δ and let P be the place of X lying under P. Then the following are equivalent: (a) P is a Weierstrass point of X of weight 5. (b) 8P is a canonical divisor on X. (c) 4P + 4Q is a canonical divisor on X. (d) 4 P is a canonical divisor on X. (e) P is a Weierstrass point of X. (f) P is a Weierstrass point of X of weight 2. 9

10 Proof: By Lemma 2.2, P has weight 5 if and only if 9 is a gap at P. The latter is equivalent to (b) since the canonical class on X has degree 8. Thus (a) and (b) are equivalent. The equivalence of (d), (e) and (f) follows similarly, using additionally Lemma 2.4(iii). The equivalence of (b) and (c) follows from Lemma 2.4(v). Each differential form on X with divisor 4 P pulls back to a differential form on X with divisor 4P + 4Q (since X is unramified over X). Hence (d) implies (c). It only remains to prove that (c) implies (d). So assume there is a differential form ω on X with divisor D := 4P +4Q. Since δ(d) = D = α 1 (D) and ω is unique up to scalar multiples, it follows that δ(ω) = ±ω and α 1 (ω) = ±ω. A non-zero regular differential form on E = X α 1 pulls back to a differential form on X with 8 distinct zeroes at the places fixed by α 1. This is the unique (up to scalar multiples) regular differential form on X fixed by α 1. Thus α 1 (ω) ω, hence α 1 (ω) = ω. Let γ := α 2 α 3 = δα 1. We are going to show that γ(ω) = ω, which implies δ(ω) = ω. The latter means that ω is the pullback of a differential form on X with divisor 4 P, completing the proof that (c) implies (d). It remains to prove that γ(ω) ω. Assume wrong. Then ω is the pullback of a differential form on X γ with divisor 4 ˆP, where ˆP is the place lying under P. Hence ˆP is a Weierstrass point of the genus 3 field X γ. Furthermore, ˆP is fixed by the bi-elliptic involution of X γ induced by α 1 (cf. proof of Lemma 2.4(iii)). This contradicts the fact that X γ is hyperelliptic (see Lemma 2.3). 3 A correspondence between certain fields of genus 3 and 5 with bi-elliptic involutions 3.1 Constructing the correspondence We fix a genus 0 field K = k(x) and consider extensions of K inside a fixed algebraic closure of K. We are interested in such extensions satisfying one of the following conditions: (*) X has genus 5 and is Galois over K such that G(X/K) is generated by 3 distinct bi-elliptic involutions α 1, α 2, α 3 of X. (**) Y has genus 3 and is Galois over K such that G(Y/K) is generated by 3 distinct bi-elliptic involutions β 1, β 2, β 3 with product 1. 10

11 Suppose first that X, α 1, α 2, α 3 satisfy (*). Define Y := X α 1α 2 α 3. Each α i induces a bi-elliptic involution β i of the genus 3 field Y by Lemma 2.4(iii). Hence Y, β 1, β 2, β 3 satisfy (**). Let B be the Klein four group generated by the involutions α i α j for i j. Then Z := X B has genus 2 by (2) and Lemma 2.3, hence Z is ramified over exactly 6 places of K. These comprise all places of K ramified in X. Since B does not contain α 1 α 2 α 3, we get that X is the compositum of Y (= X α 1α 2 α 3 ) and Z (= X B ). Conversely, consider now any Y, β 1, β 2, β 3 satisfying (**). Then Y/K is ramified at exactly 6 places of K. Define X as the compositum of Y and the unique quadratic extension Z of K ramified exactly at these 6 places of K. The three fields properly between Y and K are just the (elliptic) fixed fields of the 3 involutions in G(Y/K). Thus Z Y and it follows that X is Galois over K with elementary abelian Galois group of order 8. Since X/K is ramified at exactly the same 6 places of K as Y, it follows that X has genus 5 and is unramified over Y and over Z. Thus the involutions in G(X/Y ) and in G(X/Z) are fixed-point-free. Each β i fixes exactly 4 places of Y, which lie over 2 places P i, Q i of K. Since G(X/K) is an abelian 2-group, each place of X over P i (resp., Q i ) has the same stabilizer in G(X/K), and this stabilizer is generated by an involution α i (resp., α i). Neither α i nor α i acts trivially on Y or Z (since the involutions in G(X/Y ) and in G(X/Z) are fixed-point-free). Thus α i and α i induce the same involution of G(Y/K) (namely, β i) and the same involution of G(Z/K) (which generates G(Z/K)). Thus α i = α i fixes the 8 places of X over {P i, Q i } and is therefore a bi-elliptic involution. Thus (X, α 1, α 2, α 3 ) satisfies (*). Moreover, α 1 α 2 α 3 G(X/Y ) since β 1 β 2 β 3 = 1. We have α 1 α 2 α 3 1 by Lemma 2.4, hence Y = X α 1α 2 α 3. We have proved: Proposition 3.1 The above defines a one-to-one correspondence between the (X, α 1, α 2, α 3 ) satisfying (*) and the (Y, β 1, β 2, β 3 ) satisfying (**). If X is a Humbert field, i.e., admits two further bi-elliptic involutions α 4, α 5 with α 1 α 2 α 3 α 4 α 5 = 1, then Y = X α 1α 2 α 3 = X α 4α 5 is hyperelliptic by Lemma 2.3. The converse also holds: Remark 3.2 In the above correspondence, X is a Humbert field if and only if Y is hyperelliptic. 11

12 It remains to show that the condition is sufficient. So we assume that Y is hyperelliptic. Let β be the hyperelliptic involution of Y and V := β 1, β 2, β 3. Then β centralizes V, but β V. Hence β induces an involution γ of the genus 0 field Y V permuting the 6 places of Y V ramified in Y. Each extension of γ to the algebraic closure of Y V fixes Y as well as the unique quadratic extension Z of Y V ramified exactly at these 6 places. This proves that β extends to an automorphism α of X. Since α induces the involution γ of Y V, we have α G(X/Y V ) = α 1, α 2, α 3. Since X/Y is unramified, there are 16 places of X lying in pairs over the 8 Weierstrass points of Y. Each of these 8 pairs is fixed by α (since α induces the hyperelliptic involution of Y ). Thus α 2 fixes the 16 places in these 8 pairs. It follows that α 2 = 1 (since we know that α 2 has order 2). Furthermore, α centralizes δ := α 1 α 2 α 3 since α fixes Y = X δ. The fixed field of α, δ is the genus 0 field Y β, hence (2) implies that α is bi-elliptic. Thus X is a Humbert field. Remark 3.3 Each automorphism of X permuting α 1, α 2, α 3 induces an automorphism of Y, and each automorphism of Y permuting β 1, β 2, β 3 lifts to an automorphism of X. 3.2 The Fermat quartic corresponds to the Wiman curve The following theorem is due to [KuKo] and [KMY]. We prove a generalization in Corollary 4.7 below. Theorem 3.4 There is exactly one genus 3 field (up to isomorphism) admitting three distinct bi-elliptic involutions with product one all of whose fixed points are Weierstrass points of weight 2. This uniquely determined genus 3 field is the function field of the Fermat quartic, see Corollary 4.7 below. (We don t need it at this point). Proposition 2.12 shows how the weight of the involution fixed points behaves under the correspondence from section 3.1. Hence this correspondence turns Theorem 3.4 into the following result: Theorem 3.5 There is exactly one genus 5 field (up to isomorphism) admitting three distinct bi-elliptic involutions all of whose fixed points are Weierstrass points of weight 5. 12

13 By Proposition 2.11 and (1) we obtain the following Corollary. Corollary 3.6 (a) There is exactly one genus 5 field X (up to isomorphism) that is Galois over a genus 0 subfield L such that X is ramified over exactly three places of L with ramification indices 2,3,8. (b) There is exactly one genus 5 field X (up to isomorphism) whose automorphism group has order > 160. Remark 3.7 The genus 5 field with the unique properties from Theorem 3.5 and Corollary 3.6 is the Wiman field, the function field of the Wiman curve (see [Wiman], [KM]). This follows from Corollary 3.6(b) because the Wiman curve has exactly 192 automorphisms (see loc.cit.). In particular, the Wiman field is the unique genus 5 field with the properties from Proposition 2.11(i). We remark that Corollary 3.6(b) follows also from the (computer-generated) Table 4 of [MSSV]. Remark 3.8 Theorem 3.5 is the corrected version of Cor. 4.1 of [dc]. It includes a new proof of Th of [dc] (the main result of that paper). We remark that the proof given in [dc] is quite obscure. 4 Further discussion of the correspondence for non-hyperelliptic Y 4.1 Normalizing the canonical model of Y Let P 2 be the projective plane over k with homogeneous coordinates x 1, x 2, x 3. Each element of the group GL(3, k) induces a projectivity of P 2 (by linear action on x 1, x 2, x 3 ). Let D be the group of diagonal matrices diag(d 1, d 2, d 3 ) GL(3, k). Let P = S 3 be the group of permutation matrices in GL(3, k). Then M = DP is the group of monomial matrices. Moreover, D (resp., M) is the centralizer (resp., normalizer) in GL(3, k) of the group V = diag(1, 1, 1), diag( 1, 1, 1), diag( 1, 1, 1) 13

14 Proposition 4.1 Let Y be a non-hyperelliptic genus 3 field admitting three distinct bi-elliptic involutions β 1, β 2, β 3 with product one. Then Y is the function field of a non-singular quartic curve C of the form x x4 2 + x4 3 + a 3x 2 1 x2 2 + a 2x 2 1 x2 3 + a 1x 2 2 x2 3 = 0 (4) with a j k, where the β i are induced by the elements of V. Proof: The non-hyperelliptic genus 3 field Y is the function field of a nonsingular quartic curve C of the form F (x 1, x 2, x 3 ) = 0 (5) for a homogeneous polynomial F (x 1, x 2, x 3 ) of degree 4. Since C is the canonical model of Y, the involutions β i induce involutions β i GL(3, k). Each β i has trace -1 by Lefschetz fixed point formula (see [Br], Cor. 12.3). Thus β i is conjugate to the diagonal matrix diag(1, 1, 1). By a change of coordinates we firstly get β 1 = diag(1, 1, 1) (6) Since β 2 centralizes β 1 (and each involution in GL(2, k) is conjugate to diag(1, 1)), we can further achieve that β 2 = diag( 1, 1, 1) (7) Then β 3 = diag( 1, 1, 1) (8) Since the projectivity induced by β 1 fixes C, we get F (x 1, x 2, x 3 ) = ±F (x 1, x 2, x 3 ) If F (x 1, x 2, x 3 ) = F (x 1, x 2, x 3 ) then F ( x 1, x 2, x 3 ) = F (x 1, x 2, x 3 ) = F (x 1, x 2, x 3 ), implying that x 1 divides F contradiction (since C is nonsingular). Analogously we get F ( x 1, x 2, x 3 ) = F (x 1, x 2, x 3 ) = F (x 1, x 2, x 3 ) = F It follows that F is of the form F (x 1, x 2, x 3 ) = ax bx4 2 + cx4 3 + a 3x 2 1 x2 2 + a 2x 2 1 x2 3 + a 1x 2 2 x2 3 14

15 with a, b, c, a j k. Since C is non-singular we have abc 0. Thus we can achieve that a = b = c = 1 by a change of coordinates with a suitable element of D. This concludes the proof. Remark 4.2 The above proof shows that for given (Y, β 1, β 2, β 3 ), the invariants a j are determined up to a coordinate change in equation (4) with an element of M. For a 1 = a 2 = a 3 = 0 equation (4) defines the Fermat quartic. Let D 4 be the subgroup of D consisting of the diag(d 1, d 2, d 3 ), where the d i are fourth roots of unity. The group D 4 induces a subgroup T of the automorphism group G of the Fermat quartic such that T is the direct product of two cyclic groups of order 4. Remark 4.3 G is the semi-direct product of T and the group = S 3 induced by P. Thus the group induced by V is the unique normal Klein four subgroup of G. 4.2 Weierstrass points of C as fixed points of involutions Let C be the normalized curve from Proposition 4.1, defined by vanishing of F (x 1, x 2, x 3 ) = x x x a 3 x 2 1x a 2 x 2 1x a 1 x 2 2x 2 3 The fixed points of β 3 on P 2 are the points with x 3 = 0 plus the point (0 : 0 : 1). The latter point does not lie on C. Thus the fixed points of β 3 on C are the points (u : v : 0) with u 4 + v 4 + a 3 u 2 v 2 = 0. Clearly, uv 0, hence the fixed points of β 3 on C are the points (u : 1 : 0) with u 4 + a 3 u = 0 (9) Lemma 4.4 The point p = (u : 1 : 0) of C is a Weierstrass point if and only if a 1 + a 2 u 2 = 0. Proof: The line L through p and (0 : 0 : 1) is the only line through p that is fixed by β 3. Hence L is the tangent to C at p. It consists of all points of the form (u : 1 : x 3 ) plus the point (0 : 0 : 1). 15

16 The point p is a Weierstrass point of C if and only if it is an inflection point, which means that the tangent intersects with multiplicity 3. Thus p is a Weierstrass point if and only if the polynomial F (u, 1, x 3 ) k[x 3 ] vanishes at x 3 = 0 with multiplicity 3. Since F (u, 1, x 3 ) = x 2 3 (x2 3 + a 2 u 2 + a 1 ), the claim follows. Lemma 4.5 Let n 3 be the number of Weierstrass points of C fixed by β 3. Then n 3 {0, 2, 4}. (a) n 3 = 4 if and only if a 1 = a 2 = 0. (b) n 3 = 2 if and only if a a2 2 = a 1a 2 a 3 with a 1 a 2 0. Proof: Since C is non-singular, β 3 must have 4 fixed points on C. (Hence a 3 ±2, but we don t need this here). These 4 fixed points fall into two V-orbits of length 2. Thus n 3 {0, 2, 4}. If all fixed points of β 3 on C are Weierstrass points then by the previous Lemma we obtain a system of 2 homogeneous linear equations in a 1, a 2 with non-zero determinant. Hence (a). Assume now p = (u : 1 : 0) is a Weierstrass point of C, but we don t have a 1 = a 2 = 0. Then a 1 + a 2 u 2 = 0, hence a 1 a 2 0 and u 2 = a 1 /a 2. Now (9) yields a a2 2 = a 1a 2 a 3. The argument can also be reversed, hence (b). Corollary 4.6 Let n i be the number of Weierstrass points of C fixed by β i, and n = n 1 + n 2 + n 3. Then n equals 0,2,4 or 12. Let {i, j, l} = {1, 2, 3}. (a) n = 12 if and only if a 1 = a 2 = a 3 = 0 (Fermat quartic). (b) n = 4 = n i if and only if a j = a l = 0 and a i 0. (c) n/2 = n i = n j = 2 if and only if a 2 i = a2 j, a2 i + a2 l = a 1a 2 a 3 0. (d) n = 2 = n i if and only if a 2 j + a2 l = a 1a 2 a 3, a 2 i a2 j, a2 i a2 l, a ja l 0. Using Proposition 4.1, we conclude: Corollary 4.7 Let Y be a non-hyperelliptic genus 3 field admitting three distinct bi-elliptic involutions β 1, β 2, β 3 with product one. Let n i be the (even) number of Weierstrass points of Y fixed by β i, and n = n 1 + n 2 + n 3. Then the following holds: (i) n equals 0,2,4 or 12. (ii) n = 12 if and only if Y is the function field of the Fermat quartic and β 1, β 2, β 3 generate the unique normal Klein four subgroup of Aut(Y ). (iii) n i = 4 if and only if β i = β 2 for some β Aut(Y ) centralizing β 1, β 2, β 3 with F Y (β) = F Y (β i ). 16

17 Proof: (i) and (ii) follow from the previous Corollary and Remark 4.3. (iii) If n i = 4 then such β is induced by the projectivity multiplying x i by a fourth root of unity and fixing the other two coordinates, see Corollary 4.6(b). For the converse, assume in the set-up of Lemma 4.4 that there is a projectivity β of P 2 fixing C and p = (u : 1 : 0) with β 2 = β 3. Then β fixes the tangent L, hence the permutes the points in L C. Since L C 3, it follows that β 3 = β 2 fixes all points in L C. However, the fixed points of β 3 lie on the line x 3 = 0, and this line is different from L. Thus L C = 1, which means that p is a Weierstrass point. 4.3 The number of bi-elliptic Weierstrass points of weight 5 of a genus 5 field Proposition 2.12 shows how the weight of the involution fixed points behaves under the correspondence from section 3.1. Hence this correspondence turns Corollary 4.7 into the following result: Theorem 4.8 Let X be a genus 5 field admitting exactly three distinct bielliptic involutions α 1, α 2, α 3. Let N i be the number of Weierstrass points of X of weight 5 fixed by α i, and N = N 1 + N 2 + N 3. Thus N is the number of bi-elliptic Weierstrass points of X of weight 5. We have: (i) N equals 0,4,8 or 24. (ii) N = 24 if and only if X is the Wiman field. (iii) N i equals 0,4 or 8. We have N i = 8 if and only if α i = α 2 for some α Aut(X) centralizing α 1, α 2, α 3 with F X (α) = 4. Proof: Let (Y, β 1, β 2, β 3 ) correspond to (X, α 1, α 2, α 3 ). Then Y = X δ, where δ = α 1 α 2 α 3. Only for claim (iii) we need a further argument. (iii) Assume N 1 = 8. Then n 1 = 4 in Corollary 4.7(iii), hence there is respective β Aut(Y ). This β lifts to an automorphism α of X by Remark 3.3. Then either α 2 = α 1 or α 2 = α 1 δ = α 2 α 3. In either case, α has order 4. Since β fixes all four points of F Y (β 1 ), we get that α leaves each of the four δ-orbits on F X (α 1 ) invariant. Hence α 2 fixes each point of F X (α 1 ), thus α 2 α 2 α 3 (c.f. Lemma 2.3). Thus α 2 = α 1. The field X α = Y β has genus 0 and no place of X outside F X (α 1 ) is ramified over X α, hence F X (α) = 4. For the converse note that α induces β Aut(Y ) of order 4 with X α = Y β of genus 0, hence F Y (β) = 4. 17

18 Remark 4.9 We obtain an example of the case N = N i = 8 in Theorem 4.8 if we let (X, α 1, α 2, α 3 ) correspond to the function field Y of the Fermat quartic, together with the bi-elliptic involutions in a non-normal Klein four subgroup V of Aut(Y ). This follows from the fact that each such V intersects the normal Klein four subgroup (cf. Remark 4.3) in a group of order 2. 5 Genus 5 curves whose automorphism group is transitive on the Weierstrass points In this section we assume that X has genus g = Genus 5 curves with A 5 -action Proposition 5.1 There is exactly one genus 5 field (up to isomorphism) that admits a faithful action of A 5 (=the simple group of order 60). This field is hyperelliptic. Proof: It is well-known that the group PGL(2, k) has a unique conjugacy class of A 5 -subgroups, and each such A 5 -subgroup has a unique orbit of length 12 on the projective line over k (consisting of the fixed points of the 5-elements of A 5 ). Hence the quadratic extension X 0 of k(x) ramified at the 12 points of such an orbit is a hyperelliptic field of genus 5 that is uniquely determined up to isomorphism (cf. Remark 2.5). Assume now that G = A 5 is a group of automorphisms of X. Then G is large, hence X G has genus 0 and in the terminology of section 1.2 we have r {3, 4}. Since G has no elements of non-prime order, it follows by Remark 2.10 that 3 and 5 occur among the d i s. Hence r = 3 and {d 1, d 2, d 3 } = {3, 3, 5} by (1). G has one conjugacy class C 3 of elements of order 3 and two classes C 5, C 5 of elements of order 5. Both (C 3, C 3, C 5 ) and (C 3, C 3, C 5 ) form a rigid tuple in G in the sense of [V1], Def. 2.15, and these two tuples are interchanged by the outer automorphism of A 5 (induced by S 5 ). Hence by [V1], Th. 2.17, there is a unique Galois extension of the genus 0 field X G (inside a fixed algebraic closure of X G ) branched at the three places of X G that ramify in X, and with corresponding ramification indices 3, 3, 5. Thus X is uniquely determined up to isomorphism by the condition that it admits a faithful 18

19 action of A 5. Since the above field X 0 has the same property, it follows that X = X 0. Remark 5.2 In the situation of the Proposition, the group Aut(X) acts transitively on W(X) and is the direct product of A 5 and the group of order 2 generated by the hyperelliptic involution (cf. [KuKi]). 5.2 Transitive action on Weierstrass points The hyperelliptic curves of any genus with transitive group action on Weierstrass points are easy to classify, see [LS]. Thus we only consider the nonhyperelliptic case. Theorem 5.3 There are exactly two non-hyperelliptic genus 5 fields (up to isomorphism) whose automorphism group acts transitively on the Weierstrass points: The Humbert field with 160 automorphisms and the Wiman field. Proof: If X is the Humbert field with 160 automorphisms or the Wiman field, then Aut(X) is transitive on W(X) by Proposition For the converse, assume now that X is non-hyperelliptic and G Aut(X) acts transitively on W := W(X). Then all Weierstrass points have the same weight w, where w 5 by [Ka1]. If w = 5 then X is the Wiman field by [KM]. Hence we may assume w 4. Then W = 120 is divisible by w 5. Thus G has an element τ of order 5. This τ has exactly two fixed points P, P by Remark Hence P, P W. Furthermore, the following Claim 1 follows: Claim 1: The G-orbit B of P satisfies B 1 mod 5 or B 2 mod 5. Claim 2: The order of G P is not divisible by 2 or 3. Otherwise G P contains an element σ of order 2 or 3. Then F X (σ) 4 by Remark Since τ acts on F X (σ), but cannot fix all points of F X (σ), it follows that F X (σ) 5. Thus F X (σ) W by Schoeneberg s Lemma (see [Sch], [Le], [MV]). Hence P W, contradiction. Claim 3: G P = τ. Otherwise G P has order 25 by Claim 2, but a genus 5 field has no automorphism of order 25: By Wiman s bound (see [Br], Cor. 9.6) the order of each automorphism of X g is 4g

20 Claim 4: G is not a multiple of 30. Assume wrong. Then G = 30d, where 1 d 4, by Proposition Thus B = 6d by Claim 3. Now Claim 1 yields d 6d 1 mod 5 or d 6d 2 mod 5. If d = 1 then τ has exactly one fixed point on B, hence G acts as a Frobenius group on B contradiction, since there is no Frobenius group of permutation degree 6. It remains to consider the case d = 2, i.e., G = 60. By Proposition 5.1 we have G A 5. Hence G is solvable, thus its commutator subgroup G is a proper subgroup. If [G : G ] = 5 then G is a group of order 12 with an automorphism of order 5. Such a group does not exist. Thus [G : G ] 5. Hence G has a normal subgroup H of index 2 or 3. This H contains τ, hence the H-orbit of P has length 6 or 4. The latter case cannot occur because F X (τ) = 2. In the former case we get a contradiction as in the case d = 1. This proves Claim 4. Claim 5: X is the Humbert field with 160 automorphisms. Proof: If w 2 or w = 4 then W, hence also G is a multiple of 30. This cannot occur by Claim 4. Thus w = 3, hence W = 40. Then G is a multiple of 40. If G > 120 then X is the Humbert field with 160 automorphisms by Proposition By Claim 4 we have G 120. If G = 40 then B = 8, contradicting Claim 1. The only remaining case is G = 80 and B = 16. Then τ fixes exactly one point of B. Since W = 40, there is an involution α G fixing a point of W. We can now conclude as in the proof of Proposition 2.11 that α is bielliptic. Thus each point of W is fixed by a bi-elliptic involution, and these 5 bi-elliptic involutions are permuted transitively by τ. Thus by Lemma 2.6(iii), X is the Humbert field with 160 automorphisms. References [ACGH] E. Arbarello, M. Cornalba, P.A. Griffiths, J. Harris, Geometry of Algebraic Curves I, Springer Grundlehren 167, 1985 [Acc1] R. Accola, Riemann surfaces, theta functions and abelian automorphism groups, Lect. Notes in Math. 483 Springer Verlag [Acc2] R. Accola, Topics in the theory of Riemann surfaces, Lect. Notes in Math. 1595, Springer Verlag

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